1
$\begingroup$

I am trying to prove that multiplicity of an eigenvaliue $\lambda$ = $\dim V_{\lambda}$ and I have problems with this inequality:

$\dim V_{\lambda} \le $ multiplicity $\lambda$.

I know that characyeristic polynomial of $W_f(t)=f|_{V_{\lambda}}$ (where $f$ has Jordan form) is $\pm(t-\lambda)^{dimV_{\lambda}}$ because there are only $\lambda$'s on the matrix's diagonal.

I also need to prove that $\pm(t-\lambda)^{dimV_{\lambda}}$ divides $W_f(t)$, but I don't know how. Could you help me with that?

$V_{\lambda} = \bigcup_{k=0} ^{\infty} V_{\lambda} ^{k}$, $V_{\lambda} ^{k} = ker((f-\lambda)^k)$

By multiplicity I mean algebraic multiplicity.

$\endgroup$
  • $\begingroup$ What exactly do you mean by multiplicity of an eigenvalue, and what by $V_\lambda$? For example the multiplicity could refer to the power in the characteristic or the minimal polynomial. $V_\lambda$ could be the eigenspace or the generalised eigenspace. $\endgroup$ – Simon Markett May 22 '13 at 9:08
  • $\begingroup$ I've already corrected my question. $\endgroup$ – Sandy May 22 '13 at 9:16
  • $\begingroup$ $V_{\lambda}$ should be given by a sum, not a union. $\endgroup$ – Christopher A. Wong May 22 '13 at 9:50
  • $\begingroup$ What do you mean by that? $\endgroup$ – Sandy May 22 '13 at 9:56
  • $\begingroup$ @ChristopherA.Wong, the $V_\lambda^k$ are nested, so the union is fine. $\endgroup$ – Simon Markett May 22 '13 at 10:10
1
$\begingroup$

Hint: Over the complex numbers you have both,

$$\sum_\lambda \dim V_\lambda=n$$

and

$$\sum_\lambda mult(\lambda)=n$$

where $n$ is the size of the matrix.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm sorry, but I still don't see how to use it to prove that algebraic multiplicity is equal to geometric multiplicity. $\endgroup$ – Sandy May 22 '13 at 14:49
  • $\begingroup$ Didn't you say that you already showed the inequality in one direction? If you have $mult(\lambda)\leq dim V_\lambda$ for all eigenvalues, but the sum has to be equal... $\endgroup$ – Simon Markett May 22 '13 at 14:53
  • $\begingroup$ Ok, I've already figured that out. Thanks. $\endgroup$ – Sandy May 22 '13 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.